Show that there is a unique $r$ in real separates $A$ and $B$, such that $(-\infty, r)\subset A$ and $(r,\infty)\subset B$.

Question

Let $A, B$ be two nonempty subsets of $\mathbb{R}$, such that the union of the two is $\mathbb{R}$. Assume that any element in $A$ is strictly less than any element in $B$. Show that there is a unique $r$ in real separates $A$ and $ B$, such that $(-\infty, r)\subset A$ and $(r,\infty)\subset B$.

Proof:

1. It's not hard to show that $A\cap B=\varnothing$ by assuming not empty then contradiction.

2. Since $A$ is bounded above by any element in $B$, least upper bound exists, call it $u$. It's not hard to show that $(-\infty, u)\subset A$. Proof by contradiction.

3. Similarly, there is a greatest lower bound for the set $B$, call it $l$.

4. Note the union of these two sets is the entire real, so in order to prove the uniqueness of such a separating point, we can assume WLOG $l<u$ (proof by contradiction, then they have to be equal.) since there exists some element between those two boundaries and contained in both sets which contradicts the empty intersection, so we have $l=u$. Uniqueness is proved.

Is this correct?

Answer 1

The first three steps of your proof (as an outline) are fine but $4$ seems to fall apart. You say "so in order to prove the uniqueness of such a separating point" but you haven't proven an existence of such a point. You haven't proven that either $u$ or $l$ are such a point. By proving $l = u$ that will prove that $l = u = r$ would be such a point but not that it is unique. You can't assume wlog that $l < u$. You can prove that if $l > u$ leads to a contradiction (as $u < x < l \implies x \not \in A \cup B$) and that $u > 1$ leads to a contradiction ($l < x <u \implies \exists a\in A; b\in B: b < x < a$). That would prove $l=u$ which in turn proves $(\infty, r=u)\subset A, (r=l, \infty)\subset B$ is such a dividing $r$. But it doesn't prove it is unique.

Maybe what you want to say first is that i) if there exists such an $r$ so that $(-\infty, r) \subset A$ and $(r, \infty)\subset B$ that $r$ must be equal to both the least upper bound of $A$ and then greatest lower bound of $B$. Thus if such a point exists it must be unique. ii) Likewise if $\sup A = \inf B$ then $r=\sup A = \inf B$ would be such a point. iii) And then prove $\sup A = \inf B$.

Answer 2

hint

As you said, it is not difficult to show that

$$\forall (a,b)\in A \times B$$ $$a\le sup A\le inf B \le b .$$

and $$\sup A= \inf B= r $$ else $$c=\frac {\sup A+\inf B}{2} \in \Bbb R$$ and $$c\notin A\cup B. $$

now for any given $\epsilon>0$,

$$\exists (\alpha,\beta)\in A\times B :$$

$$r-\epsilon<\alpha <r <\beta <r+\epsilon $$ thus $r-\epsilon \in A $ and $r+\epsilon \in B $

which means that $(-\infty,r)\subset A\;\;$ and $ (r,+\infty)\subset B .$